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\title{Summary of Laplace transforms}
\date{\today}

\begin{document}
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\ 

\vspace{-.1in}

\maketitle

\vspace{-.2in}

\begin{section}{Methods of solution}

\mtd{
To solve an IVP:
\begin{enumerate}[1)]
\item Take the Laplace transform of the differential equation;
\item Use the initial conditions;
\item Solve for $Y = L(y)$ (do as little algebra as possible);
\item Take the inverse Laplace transform to find the solution.
\end{enumerate}
}

\vfill

\mtd{
To solve an IVP of the form $a y'' + by' + cy = f(t) + \alpha\delta(t - t_0)$, $y(0) = k_0$, $y'(0) = k_1$:
\begin{enumerate}[1)]
\item Find a solution $\hat{y}(t)$ to the IVP   $a y'' + by' + cy = f(t)$, $y(0) = k_0$, $y'(0) = k_1$ (use any method);
\item Find the impulse response $w(t)$ of $a y'' + by' + cy = \delta(t - t_0)$ (definition below);
\item The solution to the original IVP is $\displaystyle \boxed{y(t) = \hat{y}(t) + \alpha u(t-t_0) w(t-t_0)}$.
\end{enumerate}
}

\vfill

\thm{
Let $t_0 \geq 0$.
The solution to the IVP $a y'' + by' + cy = \delta(t - t_0)$, $y(0) = 0$, $y'(0) = 0$, is 
$\displaystyle \boxed{ y = u(t-t_0) w(t-t_0)}$ where $\displaystyle \boxed{ w(t) = L^{-1}\left( \frac{1}{as^2 + bs + c} \right)}$ ($w(t)$ is the \textbf{impulse response} of the system). 
}

\end{section}


\begin{section}{Theorems for Laplace transforms}

\thm{
Suppose $f$ and $f'$ are continuous on $[0, \infty)$ and of exponential order $s_0$, and that $f''$ is piecewise continuous on $[0,\infty)$. 
Then $f$, $f'$, and $f''$ have Laplace trannsforms for $s> s_0$:
\[ \boxed{L(f') = s L(f) - f(0)} \text{ and } \boxed{L(f'') = s^2 L(f) - sf(0) - f'(0)} \]
}


\thm{[Linearity]
Let $c_1, c_2, \dots, c_n$ be constants and let $f_1, f_2, \dots, f_n$ and $F_1, F_2, \dots, F_n$ be functions.
Then \[ L(c_1 f_1 + c_2 f_2 + \dots + c_n f_n) = c_1 L(f_1) + c_2 L(f_2) + \dots + c_n L(f_n) \]
and
\[ L^{-1}(c_1 F_1 + c_2 F_2 + \dots + c_n F_n) = c_1 L^{-1}(F_1) + c_2 L^{-1}(F_2) + \dots + c_n L^{-1}(F_n) \]
}

\thm{[Superposition]
If $y_{1}$ is the solution of the IVP \[
ay'' + b y' + c y = f_1(t),\ y(0)= k_1,\ y'(0) = k_2
\] and $y_{2}$ is the solution of the IVP \[
ay'' + b y' + c y = f_2(t),\ y(0) = l_1,\ y'(0) = l_2,
\] then $y_1+y_2$ is the solution to the IVP \[ 
ay'' + b y' + c y = f_1(t) + f_2(t),\ y(0) = k_1 + l_1,\ y'(0) = k_2 + l_2.
\]
}

\thm{[First Shifting Theorem]
If $\displaystyle L\left( f(t) \right) = F(s)$ for $s > s_0$, then $\displaystyle \boxed{ L\left( e^{at}f(t) \right) = F(s-a)}$ for $s > s_0 +a $.
}

\thm{[Second shifting theorem]
$\displaystyle \boxed{  L^{-1}\left( e^{-st_0}L \left(g(t)\right)\right) = u(t-t_0)g(t-t_0) }$ 
}

\thm{
If $t_0 \geq 0$ and $L\left( g(t + t_0)\right)$ exists for $s > s_0$, then $\displaystyle \boxed{ L\left( u(t-t_0)g(t) \right) = e^{-st_0}L \left(g(t+t_0)\right)}$ for $s> s_0$.
}

\thm{[Convolution Theorem]
$\displaystyle \boxed{L(f*g) = L(f) L(g) }$
}

\end{section}

\newpage

\begin{section}{Special functions and operations}

\dfn{
The \textbf{unit step function} (or \textbf{Heaviside function}) is $\displaystyle u(t) = \begin{cases}
0 &\text{ if } t < 0 \\
1 &\text{ if } t \geq 0
 \end{cases} $
}

\mtd{
The piecewise function $\displaystyle 
f(t) = \begin{cases} g(t) &\text{ if } t < t_1 \\ h(t) &\text{ if } t \geq t_1
\end{cases} $
can be expressed as $\displaystyle f(t) = g(t) + u(t-t_1)\left[h(t) - g(t) \right]$
}

\dfn{
Let $f$ and $g$ be functions such that if $t<0$, then $f(t) = 0$ and $g(t) = 0$.
The \textbf{convolution} of $f$ and $g$ is the function $f * g$ defined by \[ (f*g)(t) = \int_0^t f(\tau)g(t-\tau) d\tau \]
}

\dfn{
The \textbf{Dirac delta} (or \textbf{unit impulse}) function is defined to be the ``function'' $\delta(t)$ such that
\begin{enumerate}
\item $\delta(t) = 0$ if $t\neq 0$
\item $\displaystyle \int_{\infty}^{\infty} \delta(t) dt = 1$
\end{enumerate}
Note that there is no real number $y$ such that $\delta(0)= y$. 
}

\dfn{
The \textbf{Laplace transform} of $f(t)$ is $\int_0^{\infty} e^{-st}f(t)dt$.
Note, however, that this is rarely used.
Instead, refer to the a table of Laplace transforms (see section 8.8 of the textbook).
}

\end{section}

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